Let $f$ be a modular form of level $N$ and real character $\chi$ of mod $N$ and weight $k$. Does the Fourier coefficient or hecke-eigenvalue of $f$ have to be real?
What I knew is that if $N=1$ and $\chi$ trivial then fourier coefficents are real.
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$\begingroup$ Aren't Hecke operators always Hermitian with respect to Petersson inner product? $\endgroup$– Subhajit JanaCommented Apr 23, 2014 at 2:57
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$\begingroup$ @Kunnysan: Hecke operators are normal, but not always Hermitian. $\endgroup$– GH from MOCommented Apr 23, 2014 at 15:06
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2 Answers
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When acting on forms with trivial character, the Hecke operators $T_n$ for $n$ coprime to $N$ are Hermitian, so their eigenvalues are always real. This doesn't work for the Hecke operators of index not coprime to $N$; but if you have an eigenform $f$ which is new of level $N$, then its Hecke eigenvalues are all real (since the ones for $p \nmid N$ determine the rest, by the strong multiplicity one theorem). In particular, if $f$ is normalized, then its Fourier coefficients are real too (because they equal the Hecke eigenvalues).
For an explicit example where the eigenvalues are not real: take your favourite cuspidal eigenform $f$ of level $N$ and a prime $p$ not dividing $N$; then there will be two eigenforms of level $pN$ with the same eigenvalues as $f$ away from $p$, and their Hecke eigenvalues at $p$ are the roots of $X^2 - a_p(f) X + p^{k - 1}$, which are never real.
answered Apr 23, 2014 at 6:56
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2$\begingroup$ In the case of non-trivial character and $\gcd(n,N) = 1$, then the $n$th coefficient $\lambda(n)$ of a newform satisfies $\lambda(n) = \chi(n) \overline{\lambda(n)}$. It is still possible to have real coefficients if $\chi \ne 1$, but this only happens when $\lambda(n) = 0$ for all $n$ with $\chi(n) = -1$. $\endgroup$ Commented Apr 23, 2014 at 12:06
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1$\begingroup$ Jeremy, you should transform your comment into an answer: it really answers by the negative the original question. David's answer concerns only the case of trivial character. (Even better if you can add a concrete example). $\endgroup$– JoëlCommented Apr 23, 2014 at 13:21
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$\begingroup$ @Jeremy: I agree that you should turn your comment into an answer, and perhaps also elaborate on CM forms. $\endgroup$ Commented Apr 23, 2014 at 13:31
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I was encouraged to turn my comment into an answer. Theorem 6.20 of Iwaniec's book "Topics in Classical Automorphic Forms" states that if $f(z) = \sum \lambda(n) q^{n}$ is a newform (a normalized Hecke eigenform that is in the new subspace), then $$ \lambda(n) = \chi(n) \overline{\lambda(n)} $$ and so $\lambda(n)$ is real if $\chi(n) = 1$, while if $\chi(n) \ne 1$ then either $\lambda(n)$ is not real, or $\lambda(n) = 0$.
For example, the space of cusp forms of weight $4$ for $\Gamma_{0}(8)$ with character $\chi_{2}(n) = \left( \frac{2}{n}\right)$ has dimension $2$ and one of the newforms is $$ q + (-1 - \sqrt{-7}) q^{2} + 2 \sqrt{-7} q^{3} + (-6 + 2 \sqrt{-7}) q^{4} - 4 \sqrt{-7} q^{5} + \cdots $$ and the other is the complex conjugate.
It can happen that $\lambda(n)$ is real for all $n$ even if $\chi \ne 1$. For example, $$ f(z) = q \prod_{n=1}^{\infty} (1-q^{n})^{3} (1-q^{7n})^{3} $$ is a newform of weight $3$, level $7$ with character $\chi = \left(\frac{n}{7}\right)$. Such forms must be "CM forms" and arise from a Hecke grossencharacter. (See Theorem 12.5 of Iwaniec's book for details.)
answered Apr 23, 2014 at 14:57